2025/26 AHSMC Part 1 Problem 15

Let $S = \{s_1 < s_2 < s_3 < s_4 < s_5\}$. For $2 \le j \le 5$, comparing $3$-element subset averages shows $s_j - s_1$ is divisible by $3$, and comparing $4$-element subset averages shows $s_j - s_1$ is divisible by $4$. Hence $12 \mid (s_j - s_1)$ for each $j \ge 2$, so $s_5 \ge 1+12\cdot 4 = 49$. The set $\{1,13,25,37,49\}$ has all subset averages integers (any $k$-element sum with $2\le k\le 4$ is $k$ plus a multiple of $12$; the full average is $25$). Thus the minimal possible largest element is $49$. The answer is (D).